How to find the **equation of a circle** is an important concept in the realm of **geometry**. Embarking on exploring the elegance of **geometry**, this article will delve into the details of the circle. **Circles** are everywhere, from the celestial bodies in the sky to the wheels on which our cars run, making understanding their mathematical representation indispensable.

In this article, we’ll explore the methods and strategies for deriving the **equation of a circle**, a powerful tool in both **pure** and **applied mathematics**.

From simple geometric relations to complex applications, we’ll illustrate how the coordinates of the **center** and the length of the **radius** can define a circle’s equation. Whether you’re a **mathematics enthusiast**, a **curious student**, or an **educator** seeking clarity, we invite you on this intriguing journey into the world of **circular reasoning**.

## How to Find the Equation of a Circle

To **find the equation of a circle** with center **$(h,k)$** and radius **$r$**, we use the following formula:

(x – h)² + (y – k)² = r²

Where:

**(h, k)**is the**center**of the circle.**r**is the**radius**of the circle.

To find the **equation of a circle**, you need to know the **center** and the **radius**. If you know the coordinates of the **center** (h, k) and the **radius** (r), you substitute these values into the equation.

However, if you are given different information, such as the **coordinates** of points on the **circle**, you may need to use these points first to determine the **center** and **radius**. For instance, if you are given three points on the **circle**, you can use them to find the equation of the circle through methods involving **distances** and **perpendicular bisectors**.

Below we present a generic representation of the circle in figure-1.

Figure-1.

In another case, if the** circle equation** is given in the general form **Ax² + By² + Cx + Dy + E = 0**, you may need to complete the **square** to transform it to the **standard form**.

Remember that, in the context of the equation, **x,** and **y** represent any point on the circle, **h** and **k** represent the circle’s **center**, and **r** represents the **radius**. This equation **encapsulates** the definition of a **circle** as the set of all points a fixed distance** (the radius)** from a given point **(the center)**.

**Properties**

The **equation of a circle** is fundamental to understanding its properties. The equation itself is based on the definition of a circle: a set of points that are **equidistant** (the radius) from a **fixed point** (the center).

Let’s explore the properties of the circle and how they relate to its equation:

**The Center**

The **center** of the **circle** is given by the point **(h, k)** in the standard equation of a circle, **(x – h)² + (y – k)² = r²**. The coordinates **h** and **k** can be any **real numbers**. The center point can be found directly from the equation in this **standard form**.

**The Radius**

The value **r** in the standard equation gives the circle’s **radius**. It is the constant distance from the **center** to any point on the circle. Like the **center**, the radius can be found directly from the standard equation of a circle. Note that the radius must be a** positive real number**.

**Points on the Circle**

Any point** (x, y)** that satisfies the equation **(x – h)² + (y – k)² = r²** lies on the** circle**. These points can be found by substituting** x** or** y** values into the** equation** and solving for the corresponding **y** or **x** values.

**Completing the Square**

If a **circle equation** is given in the general form, **Ax² + By² + Cx + Dy + E = 0**, it can be converted into standard form by a process known as **completing the square**. This process rearranges and simplifies the equation to identify the **center** (**h, k**) and the **radius** **r**.

**Diameter, Circumference, and Area**

While these properties are not directly **visible** from the **equation**, they can be calculated using the **radius**, which is part of the **equation**. The **diameter** is twice the **radius**, the **circumference** is **2πr**, and the area is **πr²**.

Remember, the **equation of a circle** provides a **roadmap** to understanding the **circle’s properties**. It is a crucial tool in **geometry** and **algebra** for describing and investigating the nature of **circles**.

**Applications **

The ability to find the **equation of a circle** has a wide range of applications across numerous fields. Here are some examples:

**Physics and Engineering**

**Circles** describe the **motion** of objects in **circular paths** or **orbits**, such as **planets**, **electrons** around a **nucleus**, or objects in **rotational motion**. Engineers use **circle equations** in designing **circular objects** or paths, such as **wheels**, **gears**, and **roundabouts**.

**Computer Graphics and Game Design**

The equation of a circle is used to create **round objects** and effects or to calculate distances and collisions in **games**. Algorithms like the **Midpoint Circle Algorithm** use the equation of a circle to draw **circular paths** on the **pixel grid** of a **screen**.

**Geography and GPS Technology**

The concept of **‘circles of latitude’** describes the division of the Earth. In **GPS technology**, the equation of a circle (or sphere, in three dimensions) is used in **trilateration** to calculate a **user’s location** from the signals of **multiple satellites**.

**Mathematics and Education**

The equation of a circle is indeed a fundamental concept in **geometry**, **algebra**, and **trigonometry**. It is a basis for understanding and applying various mathematical concepts, including the **Pythagorean theorem**, **functions**, and **complex numbers**. By exploring the **equation of a circle**, students can develop a deeper understanding of these **mathematical principles** and their **interconnectedness**.

**Astronomy**

The **orbits** of **celestial bodies** are often** approximated** as **circles** (or **ellipses**, which are related). For instance, the **transit method** for detecting exoplanets involves observing the dip in a star’s brightness as a planet** transits** in front of it, which relies on understanding the **planet’s circular path**.

**Architecture and Design**

Circles are widely used in **design** due to their **aesthetic** appeal and **symmetry**. The ability to calculate the **equation of a circle** can help in creating accurate **designs** and **models.**

**Exercise **

**Example 1**

For a **circle** with a center at **(2, -3)** and a radius of **4**, find the **equation of the circle**.

Figure-2.

### Solution

Substitute h = 2, k = -3, and r = 4 into the standard equation:

(x – 2)² + (y + 3)² = 4²

(x – 2)² + (y + 3)² = 16

**Example 2**

Compute the **equation of a circle** with a center at the origin **(0,0)** and a radius of **5**.

Figure-3.

### Solution

Substitute h = 0, k = 0, and r = 5 into the standard equation:

(x – 0)² + (y – 0)² = 5²

x² + y² = 25

**Example 3**

Compute the **equation of a circle** with a center at **(-1,2)** and a point on the circle at **(2,4)**.

### Solution

First, find the radius using the distance formula between the center and the given point:

r = √[(2 – (-1))² + (4 – 2)²]

r = √[9]

r = 3

Then substitute h = -1, k = 2, and r = 3 into the standard equation:

(x + 1)² + (y – 2)² = 3²

(x + 1)² + (y – 2)² = 9

**Example 4**

Compute the **equation of a circle** passing through the origin **(0,0)** and having the center at **(0, 4)**.

### Solution

The radius is the distance from the center to a point on the circle (the origin):

r = √[(0 – 0)² + (0 – 4)²]

r = √[16]

r = 4

Substitute h = 0, k = 4, and r = 4 into the standard equation:

x – 0)² + (y – 4)² = 4²

x² + (y – 4)² = 16

**Example 5**

Given the equation,** x² + y² – 6x + 8y – 9 = 0**, convert it to the standard form of a circle and find the** center** and **radius**.

### Solution

We can reorganize and complete the square:

x² – 6x + y² + 8y = 9

(x – 3)² – 9 + (y + 4)² – 16 = 9

(x – 3)² + (y + 4)² = 36

So, the center is at **(3, -4),** and the radius is **√36 = 6**.

**Example 6**

Compute the **equation of a circle** with diameter endpoints at **(2, 4)** and **(6, 8)**.

### Solution

First, find the center by taking the midpoint of the endpoints:

h = (2 + 6)/2

h = 4

k = (4 + 8)/2

k = 6

Then, find the radius, which is half the length of the diameter:

r = √[(6 – 2)² + (8 – 4)²]/2

r = √[16]

r = 4

Substitute h = 4, k = 6, and r = 4 into the standard equation:

(x – 4)² + (y – 6)² = 4²

(x – 4)² + (y – 6)² = 16

**Example 7**

Compute the **equation of a circle** that touches the **x-axis** at the origin **(0,0)** and passes through the point **(1,1)**.

### Solution

Since the circle touches the x-axis at the origin, the center must be of the form (0, r). The radius r is the distance from the center to the point on the circle (1,1):

r = √[(1 – 0)² + (1 – r)²]

Solving the equation r² = 1 + 1 – 2r gives:

r = 1

Substitute h = 0, k = 1, and r = 1 into the standard equation:

(x – 0)² + (y – 1)² = 1²

x² + (y – 1)² = 1

**Example 8**

Given the equation, **2x² + 2y² – 8x + 6y – 1 = 0**, convert it to the standard form of a circle and find the **center** and **radius**.

### Solution

Divide through by 2 and reorganize to complete the square:

x² – 4x + y² + 3y

= 0.5 (x – 2)² – 4 + (y + 1.5)² – 2.25

= 0.5 (x – 2)² + (y + 1.5)²

= 5.75

So, the center is at (2, -1.5), and the radius is **√5.75 ≈ 2.4**.

*All images were created with GeoGebra.*